类型1:匹配子字符串在母串中第几个位置开始出现。
Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.
Input
The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
Output
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
Sample Input
2 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 1 3 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 2 1
Sample Output
6
-1
类型1:匹配子字符串在母串中第几个位置开始出现
#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;
const int M=1e6+10;
int n,m;
int a[M],b[M];
int nxt[M];
void kmp(){//子串最长前后缀
nxt[0]=0;
int j=0,i=1;
while(i<m){
if(b[j]==b[i]){
nxt[i]=++j;
i++;
}
else if(j==0){
i++;
}
else{
j=nxt[j-1];
}
}
}
int getans(){
int i=0;
int j=0;
while(j<m&&i<n){
if(b[j]==a[i]){
j++;
i++;
}
else if(j==0){
i++;
}
else{
j=nxt[j-1];
}
}
if(j==m) return i-m+1;//位置
return -1;
}
int main()
{
int t;
cin>>t;
while(t--){
scanf("%d %d",&n,&m);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=0;i<n;i++) scanf("%d",&a[i]);//母串
for(int i=0;i<m;i++) scanf("%d",&b[i]);//子串
kmp();
int ans=getans();
if(ans==-1) printf("-1\n");
else printf("%d\n",ans);
}
return 0;
}